# How do you use the sum and difference identities to find the exact value of cos 15^@?

Oct 24, 2014

The special triangles, 30-60-90 and 45-45-90, allow us to evaluate sin and cos.

We leverage that information to evaluate $\cos \left(15\right)$.

$\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$

$\cos \left(60 - 45\right) = \cos \left(60\right) \cos \left(45\right) + \sin \left(60\right) \sin \left(45\right)$

$\cos \left(15\right) = \cos \left(60\right) \cos \left(45\right) + \sin \left(60\right) \sin \left(45\right)$

$\cos \left(15\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$

$\cos \left(15\right) = \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}}$

$\cos \left(15\right) = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$

$\cos \left(15\right) = \frac{1 + \sqrt{3}}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

$\cos \left(15\right) = \frac{\sqrt{2} + \sqrt{6}}{4}$

$\cos \left(15\right) = 0.9659258263$

Verify your results using a calculator. Make sure the calculator is in Degree Mode. 